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The focal lengths of the objective and eye-lens of a microscope are $1 \mathrm{~cm}$ and $5 \mathrm{~cm}$ respectively. If the magnifying power for the relaxed eye is 45 , then the length of the tube is
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$15 \mathrm{~cm}$
by using $m_{\infty}=\frac{\left(L_{\infty}-f_o-f_e\right) \cdot D}{f_o f_e}$
$\Rightarrow 45=\frac{\left(L_{\infty}-1-5\right) \times 25}{1 \times 5} \Rightarrow L_{\infty}=15 \mathrm{~cm}$
$\Rightarrow 45=\frac{\left(L_{\infty}-1-5\right) \times 25}{1 \times 5} \Rightarrow L_{\infty}=15 \mathrm{~cm}$
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